This makes no sense, is my textbook dumb or I am trying too hard?

In summary, the next two elementary row operations that should be performed in the process of solving the linear system represented by the given augmented matrix are to replace R2 by its sum with 3 times R3, and then replace R1 by its sum with -5 times R3. These steps will eliminate the variable x3 from the first two equations, bringing the system closer to being solved.
  • #1
flyingpig
2,579
1

Homework Statement



Consider this matrix as the augmented matrix of a linear system. State in words the next two elementary row operations that should be performed in the process of solving the system

[tex]\begin{bmatrix}
1 & -4 & 5 & 0 & 7 \\
0 & 1 & -3 & 0 & 6 \\
0 & 0 & 1 & 0 & 2 \\
0 & 0 & 0 & 1 & -5
\end{bmatrix}[/tex]

Solution from book

The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by its sum with 3 times R3, and then replace R1 by its sum with –5 times R3.



The Attempt at a Solution



I pretty much got it except this part R1 by its sum with –5 times R3

That doesn't even eliminate the -4 in the first row. Shouldn't it be [tex]R_{1} \mapsto 4R_{2} - 5R_{3}[/tex]
 
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  • #2
I think the book is correct. It asks for two steps, and what you are suggesting is not an elementary row operation. I agree that the third step would be to try and eliminate -4 in the first row and you would do that by taking the sum of R1 and 4*R2. But again, the book seems to ask for only two elementary row operations.
 
  • #3
Ryker said:
I think the book is correct. It asks for two steps, and what you are suggesting is not an elementary row operation. I agree that the third step would be to try and eliminate -4 in the first row and you would do that by taking the sum of R1 and 4*R2. But again, the book seems to ask for only two elementary row operations.

Yeah, I did it in two steps

Step1.[tex]R_{2} \mapsto 3R_{3} + R_{2}[/tex]
Step2. [tex]R_{1} \mapsto 4R_{2} - 5R_{3}[/tex]
 
  • #4
Like I said, the second step you're suggesting is not an elementary row operation. The following two are elementary row operations:

[tex]R_{1} \mapsto R_{1} - 5R_{3}[/tex]
[tex]R_{1} \mapsto R_{1} + 4R_{2}[/tex]

A combination of them, on the other hand, is not. So you need to go in steps, you can't add or subtract more than one (multiple of a) row when talking about elementary row operations. What you suggest is a row operation, just not an elementary one.
 
  • #5
But that doesn't solve the matrix then
 
  • #6
Well, yeah, two steps of applying elementary row operations don't solve the system, three do. But the book quote you posted in your original post says to
... state in words the next two elementary row operations that should be performed in the process of solving the system.
 
  • #7
Dumb book...thanks Ryker!
 
  • #8
No problem :smile:
 

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